Harmonious Contest

最新推荐文章于 2019-08-22 15:47:58 发布

原创最新推荐文章于 2019-08-22 15:47:58 发布 · 820 阅读

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本文介绍了一个简单的算法，用于判断三个给定的边长是否能构成合法的三角形，并进一步确定该三角形是锐角、直角还是钝角三角形。通过比较最长边的平方与其他两边平方和的关系来实现。

描述

This is a magic country

Harmonious society

And harmonious people

This is a wonderful contest

Harmonious students

And harmonious problems.

This is the most harmonious problem, and the question is as following:

Given three positive integers A, B and C (0<A,B,C<=100) which denote the length of three edges, please tell me whether they can make up a legal triangle.

输入

The first line is an integer T(T<=100) which indicates the number of test cases.

Each test case consists of three integers A,B and C in a line.

输出

For each test case please output the type of triangle（Acute triangle、Right triangle or Obtuse triangle）if A,B and C can make up a legal triangle, and output "NO" otherwise. One line per case.

样例输入

4
3 4 4
3 4 5
3 4 6
3 4 7

样例输出

Acute triangle
Right triangle
Obtuse triangle
NO

分析：这个题目就是判断三角形的类型，钝角，直角，锐角。最大边的平方和其余边的平方和比，如果大，钝角，反之，锐角，相等，直角。

code:

#include <iostream>
using namespace std;
int main()
{
   int n,a,b,c;
   cin>>n;
   while(n--)
   {
     cin>>a>>b>>c;
	 if(a+b>c && a+c>b && b+c>a)
	 {
              if(a*a+b*b==c*c || a*a+c*c==b*b || b*b+c*c==a*a)
				  cout<<"Right triangle"<<endl;
			  else if((a*a+b*b)>c*c && (a*a+c*c)>b*b && (b*b+c*c)>a*a)
				  cout<<"Acute triangle"<<endl;
			  else
				  cout<<"Obtuse triangle"<<endl;
	 }
	 else
		 cout<<"NO"<<endl;
   }
   return 0;
}